HAL Id: hal-01890232
https://hal.archives-ouvertes.fr/hal-01890232v2
Preprint submitted on 9 Oct 2018 (v2), last revised 22 Oct 2019 (v3)
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Stability of Standing Waves for a Nonlinear
Klein-Gordon Equation with Delta Potentials
Elek Csobo, François Genoud, Masahito Ohta, Julien Royer
To cite this version:
Elek Csobo, François Genoud, Masahito Ohta, Julien Royer. Stability of Standing Waves for a
Non-linear Klein-Gordon Equation with Delta Potentials. 2018. �hal-01890232v2�
WITH DELTA POTENTIALS
ELEK CSOBO, FRANC¸ OIS GENOUD, MASAHITO OHTA, AND JULIEN ROYER
Abstract. In this paper, we study local well-posedness and orbital stability of standing waves for a singularly perturbed one-dimensional nonlinear Klein–Gordon equation. We first establish local well-posedness of the Cauchy problem by a fixed point argument. Unlike the unperturbed case, a noteworthy difficulty here arises from the possible non-unitarity of the semigroup generating the corresponding linear evolution. We then show that the equation is Hamiltonian and we establish several stability/instability results for its standing waves. Our analysis relies on a detailed study of the spectral properties of the linearization of the equation, and on the well-known ‘slope condition’ for orbital stability.
1. Introduction
The purpose of this work is to initiate the study of the Cauchy problem for a singularly perturbed one-dimensional nonlinear Klein–Gordon equation, namely
utt− uxx+ m2u + γδu + iαδut− |u|p−1u = 0,
u(t, x) −−−−→
|x|→∞ 0,
(u(t), ∂tu(t))|t=0= (u0, u1),
(1.1)
where u : R × R → C, m > 0, α, γ ∈ R are parameters and p > 1 determines the strength of the nonlinearity. The two coefficients δ = δ(x) are singular perturbations both given by a Dirac mass at x = 0, often referred to as a ‘delta potential’ in the context of one-dimensional evolution equations. Such space-dependent problems are sometimes termed ‘inhomogeneous’, as they model wave propagation in inhomogeneous media.
The condition that u vanishes at spatial infinity reflects a common physical requirement of having spatially localized waves, sometimes called ‘solitons’. We shall in fact seek solutions of (1.1) in H1
(R).
We will show that the evolution generated by (1.1) admits a peculiar Hamiltonian formalism, with a symplectic structure depending on the coupling constant α ∈ R. Furthermore, this Hamiltonian system is phase invariant (if u is a solution, so is eiθ
u, for any θ ∈ R), and thus possesses standing wave solutions, of the form u(t, x) = eiωt
ϕ(x), with ω ∈ R and ϕ : R → R. The stability of standing waves in Hamiltonian systems with symmetries has attracted a lot of attention since the 1980’s. So far, this issue has been fairly well understood in homogeneous media, while in inhomogeneous media it is still a subject of intense research, both theoretically and experimentally. Inhomogeneous nonlinear dispersive equations appear in various fields of physics such as nonlinear optics, cold quantum gases (e.g. Bose–Einstein condensates), plasma physics, etc. More specifically, our interest in the present problem was initially motivated by [19], where (1.1) arises as an effective model for a superfluid Bose gas.
The nonlinear Klein–Gordon equation in homogeneous media has been extensively studied. A detailed presentation of the local and global well-posedness theory can be found in [5]. Orbital stability of standing wave solutions was first addressed in the classical papers of Shatah [23,24], and Shatah and Strauss [25]. They proved that, in N space dimensions, standing waves of any frequency are orbitally unstable if p > 1 + 4/N . If 1 < p < 1 + 4/N , then there exists a critical frequency ωc such that a standing wave of frequency ω is orbitally stable if ωc < |ω| < m and unstable
if |ω| < ωc. Strong instability by blow-up in finite time was studied by Liu, Ohta and Todorova [18], and by Ohta and
Todorova [20,21]. In [14], Jeanjean and Le Coz introduced a mountain-pass approach to orbital stability for the Klein– Gordon equation, which allowed them to simplify the classical proofs and to obtain new results. In [4], Bellazzini, Ghimenti and Le Coz proved the existence of multi-solitary waves for the nonlinear Klein–Gordon equation.
The effect of a singular potential on the dynamics of the nonlinear Schr¨odinger equation has recently attracted substantial attention. Well-posedness of the Cauchy problem in the presence of a delta potential was studied in [1], while scattering for this problem was addressed in [10], both analytically and numerically. The orbital stability of standing waves of the nonlinear Schr¨odinger equation with a delta potential was studied in [7–9,17] in various regimes.
Key words and phrases. Nonlinear Klein–Gordon equation, standing waves, orbital stability, delta potential.
Stability properties of so-called ‘black solitons’ (standing waves with |ϕ(x)| → 1 as |x| → ∞) were also recently addressed in [13].
The present work is a first step in the study of the nonlinear Klein–Gordon equation with delta potentials. Our main goal here is to discuss orbital stability of standing waves of (1.1). Shortly after the seminal works [23–25], a general theory of orbital stability for infinite-dimensional Hamiltonian systems with symmetries was established by Grillakis, Shatah and Strauss [11]. Their approach, based on the so-called ‘energy-momentum’ method of geometric mechanics, was recently revisited by De Bi`evre, Rota Nodari and the second author [6], and by Stuart [26]. Under general assumptions on the dynamical system, conditions are given in these papers for orbital stability and instability. Of course, in order to discuss stability of standing waves, an essential preliminary step is to prove that the Hamiltonian system under consideration is locally well-posed. We shall thus start by addressing this issue, which is far from obvious in the context of (1.1).
The singular terms in (1.1) should be interpreted in the sense of distributions. Let us assume that u and ut are
continuous at x = 0. δu is then defined by hδu, wi = Re u(0) ¯w(0), for any function w continuous at x = 0. And δut
is defined similarly. Now, solutions of the equation in (1.1) will be continuous functions satisfying the corresponding unperturbed equation (with γ = α = 0) pointwise, outside of x = 0, together with the jump condition
u0(0+) − u0(0−) = γu(0) + iαut(0). (1.2)
Formally, this relation is indeed obtained from the equation with the delta potentials by integrating it over x ∈ (−, +) and letting → 0+. The notion of solution will be made more precise in Section2, once the appropriate functional
setting has been introduced.
Although writing the delta potentials explicitly may be useful for some formal calculations, we now introduce a functional-analytic formulation, based on the jump condition (1.2), which will make our analysis more transparent. It is convenient to reformulate the initial-value problem (1.1) as a first order system for the dependent variables (u, v) = (u, ut). We will seek solutions to (1.1) with (u, v) ∈H = H1(R) × L2(R), which we regard as a real Hilbert
space, endowed with the inner product
h(u1, v1), (u2, v2)iH = hu01, u02iL2+ hu1, u2iL2+ hv1, v2iL2,
where the real L2inner product is defined as
hu, viL2= Re
Z
R
u¯v dx. Here and henceforth,0 denotes differentiation with respect to x ∈ R.
We identify L2
(R) × L2
(R) with its dual. Then the dual H∗ of H is H−1(R) × L2
(R), and for any (ϕ, ψ) ∈ L2
(R) × L2
(R) ⊂ H−1(R) × L2
(R), the duality pairing is given by
h(ϕ, ψ), (u, v)iH∗×H = hϕ, uiL2+ hψ, viL2, (u, v) ∈H .
We shall merely write h·, ·i for h·, ·iH∗×H when no confusion is possible.
The central object in our discussion of the well-posedness of (1.1) in Section2is the generator A of the corresponding linear evolution, defined as
A = 0 IdL2 ∂2 x− m2 0 , (1.3) with domain D=(u, v) ∈ H2,∗ × H1: u0(0+) − u0(0−) = γu(0) + iαv(0) ⊂H , (1.4) where H2,∗= H1(R) ∩ H2(R \ {0}).
Note that the effect of the delta potentials is encoded in the domain of the generator.
We will show that the operator A generates a C0-semigroup onH which, remarkably, may not be a unitary group. In contrast to the classical unperturbed case, it is in general only exponentially bounded. Using Duhamel’s formula and the Banach Fixed Point Theorem, we then construct, for any initial data inH , a unique local in time solution. We also prove the blow-up alternative and continuous dependence on the initial data for this solution.
Next, standing waves, which will be our main focus, are solutions of (1.1) of the form uω(t, x) = eiωtϕω(x),
where ω ∈ R, and ϕω∈ H2,∗ is real-valued and satisfies the stationary equation
which will be interpreted as
−ϕ00+ (m2− ω2)ϕ − |ϕ|p−1ϕ = 0,
a.e. x ∈ R, (1.6)
together with the jump condition
ϕ0(0+) − ϕ0(0−) = (γ − αω)ϕ(0). (1.7)
Non-trivial localized solutions to this problem exist if and only if m2− ω2> (γ − αω)
2
4 , (1.8)
in which case they are given by the explicit formula (see Proposition 1 and Remark 1 in [17])
ϕω(x) = " (p + 1)(m2− ω2) 2 sech 2 (p − 1) √ m2− ω2 2 |x| + tanh −1− γ − αω 2√m2− ω2 !# 1 p−1 . (1.9)
In particular, there are no standing wave solutions of (1.1) when m = 0.
Definition 1.1. For any fixed m, α and γ, we shall say that ω is admissible if it satisfies the relation (1.8).
In order to reveal the Hamiltonian structure of the initial-value problem (1.1), we shall follow the notation and terminology of [6]. The Hamiltonian energy functional associated with (1.1) is given by the functional
E(u, v) = 1 2ku 0k2 L2+ m2 2 kuk 2 L2+ 1 2kvk 2 L2+ γ 2|u(0)| 2− 1 p + 1 Z R |u|p+1dx. (1.10)
We shall prove in Section3 that E is a constant of the motion. Another important quantity is conserved along the flow of the solution, namely the charge, defined as
Q(u, v) = Im Z R u¯v dx − α 2|u(0)| 2. (1.11)
We will establish in Section3that E, Q ∈ C2(H , R). Let us now introduce the symplector J : H → H∗ defined by J (u, v) = (−iαδu − v, u).
This notion, somewhat more flexible than that of a symplectic map, is introduced in [6, Sec. 6] to define Hamiltonian systems. It is noteworthy that the coupling constant α appears here in the symplectic structure itself. In this framework, the equation in (1.1) is formulated as the Hamiltonian system
J d
dtU (t) = E
0(U (t)), (1.12)
where E0 denotes the Fr´echet derivative of E. A standing wave is now a solution of the form
Uω(t, x) = eiωtΦω(x), (1.13)
where Φω:= (ϕω, iωϕω) satisfies the stationary equation
E0(Φω) + ωQ0(Φω) = 0. (1.14)
We will study the orbital stability of the standing waves (1.13), for admissible values of ω ∈ R, with respect to the
symmetry group S1 acting onH through
T (θ)(u, v) = eiθ(u, v), θ ∈ R. (1.15)
This group action leaves (1.12) invariant. The corresponding notion of orbital stability is the following.
Definition 1.2. For a fixed ω0∈ R, the standing wave eiω0tΦω0 is orbitally stable if the following holds: for any > 0
there is a δ > 0 such that, if U (t) is a solution of (3.3), then we have kU (0) − Φω0kH < δ =⇒ inf θ∈R U (t) − eiθΦω0 H < for all t ∈ R. (1.16)
Otherwise, Φω0 is said to be orbitally unstable.
In addition to orbital stability, we will also prove some linear instability results. Writing a solution U of (1.12) in the form U (t) = eiω0t(Φ
ω0+ V (t)), we have that, at first order, V satisfies the linearized equation
J d
dtV (t) = L
00
ω0(Φω0)V (t), (1.17)
Definition 1.3. The standing wave eiω0tΦ
ω0 is linearly unstable if 0 is a linearly unstable solution (in the sense of
Lyapunov) of (1.17).
In Section 4, we will carry out a stability analysis based on the energy-momentum method developed in [6,11,26]. More precisely, our proofs will make use of the well-known slope condition (also known as the ‘Vahkitov–Kolokolov criterion’), which states that the standing wave Φω0 is stable/unstable provided
d dω ω=ω 0 Q(Φω) > 0 . d dω ω=ω 0 Q(Φω) < 0, (1.18)
where the charge of the standing wave (1.13) is explicitly given by Q(Φω) = Q(ϕω, iωϕω) = −ω kϕωk2L2−
α 2|ϕω(0)|
2. (1.19)
The stability/instability of Φω in fact relies on a subtle combination of the slope condition (1.18) and suitable
spectral properties of the linearization of (1.12) (see e.g. [6, Sec. 10.3] for a detailed discussion in the context of the nonlinear Schr¨odinger equation). The spectral conditions are conveniently expressed in terms of the Lyapunov functional Lω:H → R associated with (1.12), defined by
Lω(u, v) = E(u, v) + ωQ(u, v). (1.20)
Let ˜R = diag(R, IdL2) :H → H∗, where R = −∂x2+ 1 : H1(R) → H−1(R) is the Riesz isomorphism. It follows
from the results of Sections 3 and 4 that, for any ω ∈ R, Lω ∈ C2(H , R), and that ˜R−1L00ω(Φω) : H → H is a
bounded selfadjoint operator. Let us denote by σ( ˜R−1L00ω(Φω)) ⊂ R its spectrum. The relevant spectral conditions
for stability are then formulated as follows.
(S1) There exists λω ∈ R such that σ( ˜R−1L00ω(Φω)) ∩ (−∞, 0) = {−λ2ω} and the subspace ker(L00ω(Φω) + λ2ωR) is˜
one-dimensional.
(S10) ˜R−1L00ω(Φω) has two negative eigenvalues (counted with multiplicities): either there exist λω, µω∈ R such that
σ( ˜R−1L00ω(Φω)) ∩ (−∞, 0) = {−λ2ω, −µ2ω}, λ2ω6= µ2ω, and the subspaces ker(L00ω(Φω) + λ2ωR) and ker(L˜ 00ω(Φω) +
µ2
ωR) are both one-dimensional; or there exists λ˜ ω∈ R such that σ( ˜R−1L00ω(Φω)) ∩ (−∞, 0) = {−λ2ω} and the
subspace ker(L00ω(Φω) + λ2ωR) has dimension 2.˜
(S2) ker L00ω(Φω) = span{iΦω}.
(S3) Apart from the non-positive eigenvalues, σ( ˜R−1L00ω(Φω)) is positive and bounded away from zero.
In the present context, the Cauchy problem (1.1) being locally well posed, the main results of [6,11,26] imply that, if the standing wave Φω0 satisfies (S1)–(S3), then it is orbitally stable/unstable provided (1.18) holds. In case (S1) is
replaced by (S10), we will discuss linear instability of the standing waves, by means of results obtained in [12]. We shall therefore carry out a thorough spectral analysis to see when conditions (S1)–(S3) (resp. (S10)–(S3)) are satisfied, depending on the values of the parameters. By discussing the slope condition for some values of the parameters, we will then prove various stability/instability results inH and in the subspace Hrad of radial functions.
In this analysis, we shall benefit from the explicit dependence of the solution on the parameters, but the calculations required for the slope condition are rather involved. This difficulty is reflected in the intricate form of the results we present in Section 4 and explains why we decided to focus on some regimes and refrained from attempting a comprehensive analysis. Of course, numerics might come in handy to discuss the slope condition outside the scope of our analytical results. We conclude this introduction with the following table, which captures simply what ought to be checked in order to obtain stability/instability results. The integer nω (resp. nω,rad) denotes the number of negative
eigenvalues (counted with multiplicities) of the operator ˜R−1L00ω(Φω) in H (resp. Hrad).1
nω= 1 nω= 2 and nω,rad= 1 d
dωQ(Φω) > 0 orbitally stable linearly unstable d
dωQ(Φω) < 0 orbitally unstable orbitally unstable inHrad, hence inH
2. Local well-posedness of the Cauchy problem
In this section we discuss the local well-posedness of the Cauchy problem (1.1). In order to apply the standard theory of operator semigroups, we reformulate (1.1) as a first order system onH . We consider on H the operator A
1It follows from Proposition4.9and Remark4.10that n
defined by (1.3)–(1.4). Given f : R → R we set, for U = (u, v) ∈ H , F (U ) = 0 f (u) . With f (u) = |u|p−1u and U0= (u0, u1), (1.1) can be rewritten as
(
Ut(t) − AU (t) = F (U ),
U (0) = U0.
(2.1) We will show that A generates a strongly continuous semigroup on H , which will allow us to establish the local well-posedness of (1.1).
Definition 2.1. Let T ∈ (0, ∞].
• A strong solution to (2.1) is a function U ∈ C0([0, T ), D) ∩ C1([0, T ),H ) such that (2.1) holds on [0, T ). We
say that u is a strong solution of (1.1) on [0, T ) if (u, ut) is a strong solution of (2.1).
• A weak solution of (2.1) is a function U ∈ C0([0, T ),H ) such that, for all t ∈ [0, T ), there holds
U (t) = etAU0+
Z t
0
e(t−s)AF (U (s)) ds. (2.2) We say that u is a weak solution of (1.1) on [0, T ) if (u, ut) is a weak solution of (1.1).
We begin with a lemma which ensures, in particular, that A is densely defined. Lemma 2.2. D is dense inH .
Proof. Let (u, v) ∈H . We can consider a sequence (vn) in H1such that vn(0) = 0 and vn→ v in L2. We then choose
a sequence (un) in H2 which converges to u in H1. For n ∈ N and x ∈ R, we set
ζn(x) = 1 + γ |x| 2 e −nx2 . We have ζn0(0+) = −ζn0(0−) = γ2, so (unζn)0(0+) − (unζn)0(0−) = un(0) ζn0(0 +) − ζ0 n(0−) = γ(unζn)(0).
This proves that (unζn, vn) belongs to D for all n ∈ N. Moreover,
kζn− 1kL∞ −−−−−→
n→+∞ 0 and kζ 0
nkL∞ −−−−−→
n→+∞ 0,
so kunζn− unkH1→ 0, and hence (ζnun, vn) goes to (u, v) inH .
2.1. Linear evolution in the energy space. In this subsection we show that the operator A generates a strongly continuous group onH . We know that if A is skew-adjoint then it generates a one parameter unitary group on H . Since the notion of skew-adjointness depends on the inner product, we first discuss the choice of a suitable Hilbert structure onH .
For µ > 0 we introduce on H the quadratic form defined by k(u, v)k2
H ,µ,γ = ku0k2L2+ µ2kuk2L2+ γ|u(0)|2+ kvk2L2. (2.3)
We denote by h·, ·iH ,µ,γ the corresponding bilinear form. With µ = m we observe that, for U = (u, v) ∈ D, hAU, U iH ,m,γ= hv0, u0iL2+ m
2
hv, uiL2+ γ Re(v(0)u(0)) + hu00, viL2− m
2
hu, viL2
= hv0, u0iL2+ γ Re(v(0)u(0)) − hu0, v0iL2− Re(γu(0)v(0) + iα|v(0)|
2
) = 0.
(2.4)
This makes h·, ·iH ,m,γa good candidate to be a suitable inner product onH . However, for negative γ, it may happen that the corresponding quadratic form takes negative values. In this case we have to choose a larger parameter µ. Lemma 2.3. Let µ0= ( 0 if γ > 0, |γ| 2 is γ < 0.
Then for µ > µ0 there exists Cµ> 1 such that, for all u ∈ H1, we have
Cµ−1kuk2H1 6 ku0k 2 L2+ µ 2kuk2 L2+ γ |u(0)| 2 6 Cµkuk 2 H1. (2.5)
In particular, the functional k · kH ,µ,γ is a norm onH , equivalent to the usual one. Proof. For u ∈ H1we have
|γ| |u(0)|2= 2 |γ| ReZ 0 −∞ u(x)¯u0(x) dx 6 2 |γ| kukL2ku0kL2 6 ku0k2 L2 2 + 2γ 2kuk2 L2. (2.6)
This gives in particular the second inequality of (2.5). By Theorem I.3.1.4 in [2], we have
ku0k2L2+ γ |u(0)|
2
+ µ20kuk2L2 > 0
for all u ∈ H1. Then for > 0 we have
ku0k2L2+ µ 2kuk2 L2+ γ |u(0)| 2 > 2 ku0k2L2+ 2γ |u(0)| 2 + (µ2− (1 − 2)µ2 0) kuk 2 L2 > ku0k2L2+ µ 2− (1 − 2)µ2 0− 4γ 2 kuk2 L2.
With > 0 small enough, this gives the first inequality in (2.5), and the second statement of the proposition follows. We intend to prove the following proposition.
Proposition 2.4. The operator A generates a C0-semigroup on H . Moreover, there exist M > 0 and β > 0 such
that, for all t ∈ R, we have
etA L(H )6 M eβ|t|. For this we need the following lemma.
Lemma 2.5. Let µ > 0 be as in Lemma2.3and λ >pµ2− m2.
(i) The bounded operator
−∂xx+ (m2+ λ2) + (γ + iλα)δ : H1→ H−1 (2.7)
has a bounded inverse, which we denote by R(λ).
(ii) Let ϕ ∈ L2and ψ ∈ H1. Then R(λ)(ϕ − δψ) belongs to H2,∗. It is the unique solution u in H2,∗of the problem (
−u00+ (m2+ λ2)u = ϕ,
u0(0+) − u0(0−) = (γ + iλα)u(0) + ψ(0). (2.8)
Proof. By Lemma2.3we have, for all u ∈ H1,
h(−∂xx+ (m2+ λ2) + (γ + iλα)δ)u, uiH−1,H1 > ku0k2L2+ µ2kuk2L2+ γ|u(0)|2& kuk
2 H1.
Similarly, for all u and v in H1,
h(−∂xx+ (m2+ λ2) + (γ + iλα)δ)u, viH−1,H1. kukH1kvkH1.
Hence, the operator (2.7) has a bounded inverse by the Lax–Milgram Lemma. Let us assume that (2.8) has a solution u ∈ H2,∗. For all w ∈ H1 we have
h−u00, wiL2+ (m2+ λ2)hu, wiL2 = hϕ, wiL2.
Integrating by parts and using the jump condition (2.8), we get
hu0, w0iL2+ (m2+ λ2)hu, wiL2+ Re((γ + iλα)u(0) ¯w(0)) = hϕ, wiL2− Re(ψ(0) ¯w(0)). (2.9)
This proves that u = R(λ)(ϕ − δψ). Conversely, let u = R(λ)(ϕ − δψ) ∈ H1. Then (2.9) holds for all w ∈ C0∞(R \ {0}),
so u belongs to H2,∗ and −u00+ (m2+ λ2)u = ϕ. We now write (2.9) with w ∈ C0∞(R) such that w(0) = 1, which
yields the jump condition in (2.8).
We can now prove Proposition2.4.
Proof of Proposition2.4. Consider µ as given by Lemma2.3. For U = (u, v) ∈ D, we have hAU, U iH ,µ,γ= hv0, u0iL2+ µ 2hv, ui L2+ γ Re(v(0)u(0)) + hu00, viL2− m 2hu, vi L2 = hv0, u0iL2+ (µ 2− m2) hv, ui
L2+ γ Re(v(0)u(0)) − hu0, v0iL2− Re γu(0)v(0) − iα|v(0)|
2 = (µ2− m2) hv, ui
and so |hAU, U iH ,µ,γ| = (µ2− m2)|hu, viL2| 6 µ2− m2 2 (kuk 2 L2+ kvk 2 L2).
On the other hand, by Lemma2.3,
kU k2H ,µ,γ & kuk2L2+ kvk
2 L2.
Hence, fixing β > 0 large enough, we have
h(±A − β)U, U iH ,µ,γ 6 0. (2.10)
Therefore, by [5, Proposition 2.4.2], the operators ±A − β are dissipative. In particular, for λ > β, we have
k(±A − λ)U k2H ,µ,γ > k(±A − β)U k2H ,µ,γ+ (λ − β)2kU k2H ,µ,γ, (2.11) so that ±A − λ are injective with closed range. Now, let F = (f, g) ∈H . For U = (u, v) ∈ D, we have
(A − λ)U = F ⇐⇒ (
v = λu + f,
u00− (m2+ λ2)u = g + λf.
By Lemma2.5, if β is large enough, there exists U = (u, v) ∈ D such that the right-hand side is satisfied. It is given by u = R(λ)(−g − λf − iαδf ) and v = λu + f . This proves that Ran(A − λ) =H . Hence, (A − λ) has a bounded inverse and, by (2.11), (A − λ)−1 L(H )6 1 λ − β.
By the Hille–Yosida Theorem, this proves that A generates a C0-semigroup on (H , k·k
H ,µ,γ). Furthermore, for t > 0
and U ∈H , we have
etAU H ,µ,γ 6 eβtkU kH ,µ,γ.
Since the norm k·kH ,µ,γ is equivalent to the usual one, there exists M > 1 such that we also have
etAU
H 6 M e βtkU k
H .
Now the same holds true with A replaced by −A, and the proof is complete.
Remark 2.6. If γ > −m, we can chose µ = m in Lemma2.3 and 2.5. As in the proof of Proposition 2.4, we can show that A is skew-adjoint and, by [5, Theorem 3.2.3.], it now generates a one-parameter unitary group:
etAU H ,m,γ = kU kH ,m,γ.
2.2. Local well-posedness of the nonlinear problem. We are now in a position to prove the local well-posedness of the Cauchy problem (1.1). We suppose that the general nonlinearity f ∈ C(C, C) satisfies the following:
f (0) = 0, (2.12)
|f (u) − f (v)| 6 Cf(1 + |u|p−1+ |v|p−1)|u − v|, (2.13)
where p > 1 and Cf > 0.
Lemma 2.7. For any R > 0, there exists CR > 0 such that, for u1, u2∈ H1 with ku1kH1 6 R and ku2kH1 6 R, we
have
kf (u1)kL2 6 CRku1kH1,
kf (u1) − f (u2)kL2 6 CRku1− u2kH1.
Proof. For j ∈ {1, 2} we have kujkL∞ 6 R, so with Cf we get
kf (u1) − f (u2)k 2 L2 6 Z R Cf2 1 + |u1(x)| p−1 + |u2(x)| p−12 |u1(x) − u2(x)| 2 dx 6 Cf2(1 + 2R p−1)2 ku1− u2k 2 L2.
This gives the second inequality. The first one follows by taking u2= 0.
Corollary 2.8. F :H → H is Lipschitz continuous on bounded subsets of H : for any R > 0, there is a constant L(R) such that, for U, V ∈H with kUkH 6 R and kV kH 6 R, we have
Lemma 2.9. Let T > 0, U0∈H , and U, V ∈ C([0, T ], H ) be two solutions to (2.2). Then U = V .
Proof. Let us set R = supt∈[0,T ]max{kU (t)kH , kV (t)kH}. For t ∈ [0, T ] we have kU (t) − V (t)kH 6 Z t 0 e (t−s)AF (U (s)) − F (V (s)) H ds 6 M e T βL(R) Z t 0 kU (s) − V (s)kH ds.
By Gronwall’s Lemma we conclude that U (t) = V (t) for all t ∈ [0, T ].
In the next proposition we prove the existence of a weak solution to the Cauchy problem.
Proposition 2.10. Take R > 0 and U0∈H such that kU0kH 6 R. Then there exists TR> 0 and a unique solution
U ∈ C([0, TR),H ) of problem (2.2).
Proof. We only need to prove the existence of the solution, as uniqueness follows from Lemma2.9. Let M and β be as in Proposition2.4. Consider U0∈H such that kU0kH 6 R. For T > 0 to be determined later, let
X := {U ∈ C([0, T ],H ) : kU(t)kH 6 3M R ∀t ∈ [0, T ]} and
d(U, V ) := max
t∈[0,T ]kU (t) − V (t)kH.
Then (X, d) is a complete metric space. We now define a map Ψ : X → C([0, T ],H ) by Ψ(U ) : t 7→ etAU0+
Z t
0
e(t−s)AF (U (s)) ds.
Note that, for all s ∈ [0, T ], we have kF (U (s))kH 6 3M R L(3M R) by Corollary2.8. Thus, kΨ(U )(t)kH 6 ketAU 0kH + Z t 0 ke(t−s)AF (U (s))k H ds 6 M eT βR + M eT β Z t 0 kF (U (s))kH ds 6 M eT βR + M eT β3M R L(3M R) T. (2.14)
Furthermore, for U, V ∈ E, we have
kΨ(U )(t) − Ψ(V )(t)kH 6 M eT β
Z t
0
kF (U (s)) − F (V (s))kH ds 6 M eT βL(3M R) T d(U, V ).
It is now straightforward to check that, if T = TR> 0 is chosen small enough, there holds
M eT βR+M eT β3M R L(3M R)T 6 3M R, M eT βL(3M R)T 6 1/3.
This shows that Ψ maps (X, d) to itself and is a contraction. The result now follows from the Fixed Point Theorem. Theorem 2.11. There exists a function T :H → (0, ∞] with the following properties. For all U0∈H , there exists
a function U ∈ C([0, T (U0)),H ) such that, for all 0 < T < T (U0), U is the unique solution of (2.2) in C([0, T ],H ).
Furthermore, the blow-up alternative holds: if T (U0) < ∞ then limt↑T (U0)kU (t)kH = ∞.
Proof. For all U0∈H , we set
T (U0) = sup{T > 0 : ∃U ∈ C([0, T ],H ) solution to (2.2)}.
From Proposition2.10, we know that T (U0) > TkU0k> 0 and Lemma2.9allows us to extend it to a maximal solution
U ∈ C([0, T (U0)],H ). The blow-up alternative follows from an argument by contradiction. Suppose that T (U0) < ∞
and that there exists a constant C and a sequence tn in [0, T (U0)) such that tn ↑ T (U0) and supn∈NkU (tn)kH 6 C.
Now take a time tn such that tn+ TC > T (U0). Using Lemma 2.9and Proposition2.10, we can extend the solution
up to tn+ TC by considering the initial value problem (2.2) with initial value U (tn). This contradicts the definition
of T (U0) and concludes the proof.
(i) T :H → (0, ∞] is lower semicontinuous: given the initial conditions U0, U0,n∈H such that U0,n converges to
U0 inH , we have that
T (U0) 6 lim inf
n→∞ T (U0,n).
(ii) If U0,n → U0 and if T < T (U0), then Un → U in C([0, T ],H ), where Un and U are the solutions of (2.2)
corresponding to the initial data U0,n and U0.
Proof. Let U0∈H and U ∈ C([0, T (U0)),H ) be the solution of (2.2) given by Theorem2.11. Let 0 < T < T (U0).
It suffices to show that, if U0,n→ U0 then T (U0,n) > T for n large enough, and Un → U in C([0, T ],H ). We set
R = 2 sup
t∈[0,T ]
kU (t)kH, and
τn= sup{t ∈ [0, T (U0,n)) : kUn(s)kH 6 2R ∀s ∈ [0, t]}.
If n is large enough, we have kU0,nkH 6 R. Hence, by Proposition2.10, 0 < TR< τn. Now, for all 0 < t 6 min{τn, T },
kU (t) − Un(t)kH 6 etA(U0− U0,n) H + Z t 0 e (t−s)A(F (U (s)) − F (U n(s))) H ds 6 M eT βkU0− U0,nkH + M eT βL(2R) Z T 0 kU (s) − Un(s)kH ds.
Therefore, by Gronwall’s lemma,
kU (t) − Un(t)kH 6 M eT βkU0− U0,nkH eM e
T βL(2R)T
(2.15) for all t 6 min{T, τn}. In particular, if n is large enough,
kUn(t)kH 6 R
for t 6 min{T, τn}. Hence τn > T , which implies that T (U0,n) > T . From (2.15) we also see that Un → U in
C([0, T ],H ), which completes the proof.
Theorem 2.13. Let U0∈ D and T ∈ (0, T (U0)). Let U ∈ C([0, T ],H ) be the corresponding solution of (2.2). Then
U ∈ C([0, T ], D) ∩ C1([0, T ],H ).
Proof. Let h > 0 and t ∈ [0, T − h). By a change of variables it is easy to see that U (t + h) − U (t) = e(t+h)AU0− etAU0+ Z t 0 esA F (U (t + h − s)) − F (U (t − s)) ds + Z h 0 e(t+h−s)AF (U (s)) ds. Hence, kU (t + h) − U (t)kH 6 etA(ehAU0− U0) H + Z t 0 esA F (U (t + h − s)) − F (U (t − s)) H ds + Z h 0 e (t+h−s)AF (U (s)) H ds 6 M eT β ehAU0− U0 H + M e T βL(R)Z t 0 kU (t + h − s) − U (t − s)kH ds + M eT βh sup s∈[0,T ] kF (U (s))kH . We know that ehAU0− U0= Z h 0 esAAU0ds, and so ehAU 0− U0
H 6 hM eβTkAU0kH. Applying Gronwall’s Lemma, we get
kU (t + h) − U (t)kH . h
for all 0 6 t < t + h 6 T . Hence, U : [0, T ] → H and F (U ) : [0, T ] → H are Lipschitz continuous. We conclude
Remark 2.14. It is worth noting that solutions of (1.1) may blow up in finite time. To this aim, let us consider the ordinary differential equation
v00(t) + v(t) − v3(t) = 0. (2.16)
For any fixed T > 0, this equation has the solution
v(t) = 1
tanh T − t/√2 , (2.17)
which blows up at time√2T . Now, consider (1.1) with m = 1, γ = α = 0 and p = 3, and choose the constant initial data u0= 1/ tanh (T ). By finite speed of propagation (see e.g. [5]), if u0 is smoothly truncated outside an interval of
length 2√2T + 1, the corresponding solution of (1.1) will blow up like (2.17) at time√2T . Again by finite speed of propagation, if the support of the truncated u0 is chosen far away from x = 0, then the solution u will not ‘see’ the
Dirac potentials over the time interval [0,√2T ), and will also blow up at time√2T , for any values of γ and α. 3. Hamiltonian structure
In this section we show that (1.1) is a Hamiltonian system, and we establish the relevant conservation laws, namely that the energy and the charge defined in (1.10) and (1.11) are constants of the motion. We shall use the general framework developed in [6] to study orbital stability of standing waves of infinite-dimensional Hamiltonian systems.
We start by showing that, in the terminology of [6, Sec. 6], (H , D, J ) forms an appropriate symplectic Banach triple for our problem, provided the map J :H → H∗ defined by
J (u, v) = (−iαδu − v, u) (3.1)
is a (weak) symplector, in the sense of Definition 6.2 (i) in [6], which we check now.
Lemma 3.1. The map J :H → H∗ defined by (3.1) is a symplector, that is: (i) J is a bounded linear map;
(ii) J is one-to-one;
(iii) J is anti-symmetric, in the sense that
hJ (u, v), (w, z)i = − hJ (w, z), (u, v)i , (u, v), (w, z) ∈H .
Proof. (i) Linearity is obvious and boundedness follows from the Sobolev embedding theorem through the estimate | hJ (u, v), (w, z)i | = Im αu(0) ¯w(0) − Re Z R v ¯w + Re Z R u¯z 6 |α| kukH1kwkH1+ kvkL2kwkL2+ kukL2kzkL2 . kukH1+ kvkL2 kwkH1+ kzkL2.
(ii) J is one-to-one since, clearly, J (u, v) = (0, 0) if and only if (u, v) = (0, 0).
(iii) The antisymmetry of J follows by a straightforward calculation, using that α ∈ R.
We now turn our attention to the regularity of the energy and charge functionals respectively introduced in (1.10) and (1.11). In particular, in the terminology of Definition 6.5 in [6], we show that E and Q have J -compatible derivatives, i.e. that E0(u, v), Q0(u, v) ∈ rge J for all (u, v) ∈ D. We write E, Q ∈ Dif(D, J ).
Lemma 3.2. We have that E ∈ C1(H , R) ∩ C2
(D, R) and Q ∈ C2(H , R). For (ϕ, ψ) ∈ D and (u, v), (w, z) ∈ H ,
we have
E0(ϕ, ψ) = (−ϕ00+ m2ϕ − |ϕ|p−1ϕ − iαδψ, ψ),
hE00(ϕ, ψ)(u, v), (w, z)i = Rehγu(0) ¯w(0) + Z R u0w¯0+ m2u − |ϕ|p−1u − (p − 1)|ϕ|p−3ϕ Re(ϕ¯u) ¯wdx + Z R v ¯z dxi, and for (ϕ, ψ), (u, v) ∈H
Q0(ϕ, ψ) = (−αδϕ + iψ, −iϕ), Q00(ϕ, ψ)(u, v) = (−αδu + iv, −iu). Furthermore, E0(ϕ, ψ) ∈ rge J and Q0(ϕ, ψ) ∈ rge J for all (ϕ, ψ) ∈H .
Proof. The regularity stated and the expressions obtained for the Fr´echet derivatives follow from routine verifications. Let (ϕ, ψ) ∈ D. To see that E0(ϕ, ψ) ∈ rge J , one has to find (w, z) ∈H such that
−iαδw − z = −ϕ00+ m2ϕ − |ϕ|p−1ϕ − iαδψ and w = ψ inH∗.
This yields (w, z) = (ψ, ϕ00− m2ϕ + |ϕ|p−1ϕ), which clearly belongs toH . Similarly, for (ϕ, ψ) ∈ H ,
J (w, z) = Q0(ϕ, ψ) ⇐⇒ (w, z) = −i(ϕ, ψ). (3.2)
This completes the proof.
Lemmas 3.1and 3.2show that (H , D, J ) is a suitable symplectic Banach triple for our problem, with associated Hamiltonian E. For initial conditions U0 = (u0, u1) ∈ D, the differential equation in (1.1) can indeed be written as
the Hamiltonian system (see Definition 6.6 in [6])
J d
dtU (t) = E
0(U (t)). (3.3)
Remark 3.3. The well-posedness theory in Section2shows that the domain D is stable under the flow of (3.3), so that, by Lemma3.2, E0(U (t)) indeed belongs to rge J over the lifespan of the solution.
Proposition 3.4. The energy E and the charge Q are constants of the motion for (3.3), i.e. for any U0= (u0, u1) ∈
H , E(U(t)) = E(U0) and Q(U (t)) = Q(U0), as long as the solution exists.
Proof. Following [6, Theorem 5, p. 191], one only needs to check that both E and Q Poisson-commute with E, i.e. that {E, E}(u, v) = {E, Q}(u, v) = 0 for all (u, v) ∈ D, where for any F ∈ Dif(D, J ), the Poisson bracket {E, F } is defined as
{E, F }(u, v) =E0(u, v), J−1F0(u, v) , (u, v) ∈ D. (3.4)
That {E, E}(u, v) = 0 for all (u, v) ∈ D is then a trivial consequence of the anti-symmetry of J . As for {E, Q}, using the explicit expression J−1(w, z) = (z, −w − iαδz) (or (3.2)), we have
{E, Q}(u, v) =(−u00+ m2u − |u|p−1u − iαδv, v), (−iu, −iv) = Reh
Z
R
(−u00+ m2u − |u|p−1u − iαδv)(−iu) dx + Re Z R v(−iv) dxi = Reh− i Z R u00u dx + αv(0)¯¯ u(0)i
= Re[−i¯u(0)(u(0−) − u(0+)) + α¯u(0)v(0)] = 0,
which completes the proof.
4. Stability of standing waves
Having established the well-posedness and the Hamiltonian structure of the initial-value problem (1.1), we now investigate the stability of standing waves by applying the energy-momentum method described in the introduction. The criterion for orbital stability of the standing waves (1.13) is the following.
Proposition 4.1. Suppose the standing wave eiω0tΦ
ω0(x) satisfies the spectral conditions (S1)–(S3). Then it is
orbitally stable if d dω ω=ω 0 Q(Φω) > 0,
and orbitally unstable if
d dω ω=ω 0 Q(Φω) < 0.
Let A be a selfadjoint operator that is bounded below with positive essential spectrum. We shall henceforth denote by n(A) ∈ N the number of negative eigenvalues (counted with multiplicities) of A, and we set
nω:= n ˜R−1L00ω(Φω),
for all admissible ω ∈ R. In Proposition4.1, we have nω0 = 1. For nω0 = 2, we will exhibit regimes of linear instability
Proposition 4.2. Let p(d00(ω)) = 1 if dωd ω=ω 0 Q(Φω) > 0, 0 if dωd ω=ω 0 Q(Φω) < 0.
Then the standing wave eiω0tΦ
ω0 is linearly unstable if nω0− p(d
00(ω
0)) is odd.
Corollary 4.3. Suppose the standing wave eiω0tΦ
ω0 satisfies (S1 0)–(S3) and d dω ω=ω 0 Q(Φω) > 0.
Then it is linearly unstable.
4.1. Spectral analysis. Our purpose here is to give some spectral properties (in particular the number of negative eigenvalues) of the operator ˜R−1L00ω(Φω). We will consider α, γ and ω satisfying (1.8). The quantity
β = β(ω) := γ − αω,
which appears in (1.5), will play an important role below. In view of the admissibility condition (1.8), we shall consider β ∈ (−β0, β0), where β0 := 2
√
m2− ω2. The main results of this subsection rely on the dependence on β of the key
objects entering the spectral analysis. With this in mind (and to avoid a too heavy notation) we shall relabel various quantities by β and temporarily drop the index ω. For instance — with a slight abuse of notation — we will write ϕβ
instead of ϕω, that is,
ϕβ(x) = " (p + 1)(m2− ω2) 2 sech 2 (p − 1) √ m2− ω2 2 |x| + tanh −1 − β 2√m2− ω2 !# 1 p−1 . (4.1)
One should of course keep in mind the dependence on ω. It will not be relevant for our analysis here, but will come back with full force in the next subsection. For U = (u, v) ∈H , we now let
L00βU := − u00+ m2u − ϕp−1β u − (p − 1)ϕp−1β Re(u) + βδu + iωv, v − iωu
∈H∗. (4.2)
In view of Lemma3.2, this reads L00β= L00ω(Φω). We shall also use the convenient notation ˜L00β := ˜R −1L00
β.
Let β ∈ (−β0, β0). We observe that L00β:H → H
∗ is a bounded operator and, for U, W ∈H , we have
h ˜L00βU, W iH = hL00βU, W iH∗,H = hU, L00βW iH ,H∗= hU, ˜L00βW iH, (4.3)
so ˜L00β is a bounded selfadjoint operator on H .
Instead of analysing directly the spectral properties of ˜L00β, it will be more convenient to work with the operator on L2× L2 associated to the form L00
β. More precisely, we set
Dβ:= {u ∈ H2,∗: u0(0+) − u0(0−) = βu(0)} (4.4)
and we consider on L2× L2 the operator L
β defined by D(Lβ) = Dβ× L2 and, for U = (u, v) ∈ D(Lβ),
LβU = − u00+ m2u − ϕp−1β u − (p − 1)ϕ p−1
β Re(u) + iωv, v − iωu
∈ L2× L2.
This defines a (R-linear) selfadjoint operator which shares the same relevant spectral properties as ˜L00β:
Proposition 4.4. The operator Lβ is selfadjoint and bounded from below on L2× L2, and for U ∈ Dβ× L2 we have
hLβU, U iL2×L2= hL 00 βU, U iH∗,H = h ˜L00βU, U iH. We have ker(Lβ) = ker( ˜L00β). (4.5) Moreover
inf σess(Lβ) > 0 ⇐⇒ inf σess( ˜L00β) > 0, (4.6)
and, in this case,
Proof. • For U, V ∈H , we set Q(U, V ) = hL00βU, V iH∗,H. This defines on L2× L2a symmetric bilinear form with
domain D(Q) =H . Using a trace inequality as in (2.6), we can check that Q is bounded from below and closed. We denote by T the corresponding selfadjoint operator given by the Representation Theorem (see for instance [15, VI-Theorem 2.1] for sesquilinear complex forms, the symmetric case being analogous for real bilinear forms). In particular,
D(T ) =U ∈H : V 7→ Q(U, V ) is a continuous linear functional on L2× L2 ,
and
hT U, V iL2×L2 = Q(U, V ), for all U ∈ D(T ), V ∈H . (4.8)
Since hLβU, V iL2×L2 = Q(U, V ) for all U ∈ D(T ) and V ∈H , we have D(Lβ) ⊂ D(T ) and Lβ= T on D(Lβ). Now let
U = (u, v) ∈ D(T ). Writing (4.8) with V = (w, 0) for any w ∈ C0∞(R∗) proves that u ∈ H2,∗. Then, with w ∈ C0∞(R)
such that w(0) = 1 or w(0) = i, we obtain u ∈ Dβ, so U ∈ D(Lβ). This means that Lβ = T , and the first part of the
proposition is proved.
• Let U = (u, v) ∈ ker(Lβ). In particular we have U ∈H and h˜L00βU, W iH = hL00βU, W iH∗,H = 0 for all W ∈H , so
U ∈ ker( ˜L00β). Conversely, if U ∈ ker( ˜L00β) then hL00βU, W i = h0, W iL2×L2 for all W ∈H , so U ∈ D(Lβ) and LβU = 0.
This proves (4.5).
• Now suppose that inf σess(Lβ) > 0. Since Lβ is bounded from below, it has a finite number ˜m of non-positive
eigenvalues (counted with multiplicities). We denote by ˜Θ the subspace of L2× L2 generated by the corresponding
eigenvectors, and by ˜Θ⊥the orthogonal complement of ˜Θ in L2× L2. We also set ˜Θ⊥
1 = ˜Θ⊥∩H . Since ˜Θ ⊂ D(Lβ) ⊂
H , ˜Θ and ˜Θ⊥
1 are complementary subspaces ofH .
There exists σ0> 0 such that σ(Lβ) ∩ (0, σ0) = ∅. Then, for all U ∈ D(Lβ) ∩ ˜Θ⊥, we have
h ˜L00βU, U iH = hL00βU, U iH∗,H = hLβU, U iL2×L2 > σ0kU k2L2×L2.
On the other hand, by the trace inequality there exists C > 0 such that, for all U ∈ H , hL00βU, U iH∗,H >
kU k2H
2 − C kU kL2×L2. Thus, for η ∈ (0, 1) and U ∈ D(Lβ) ∩ ˜Θ⊥1, we have
h ˜L00βU, U iH >η 2kU k 2 H − ηC kU kL2×L2+ (1 − η)σ0kU k 2 L2×L2.
For η > 0 small enough, this yields
h ˜L00βU, U iH > η 2kU k 2 H, for all U ∈ D(Lβ) ∩ ˜Θ⊥1. But D(Lβ) ∩ ˜Θ⊥1 is dense in Θ⊥1, so h ˜L00βU, U iH > η 2kU k 2 H , for all U ∈ ˜Θ⊥1.
Since ˜Θ⊥1 is of codimension m inH , the Min-Max Principle (see, e.g., Theorem XIII.1 in [22]) implies that inf σess( ˜L00β) >
0 and that ˜L00
β has at most ˜m negative eigenvalues (counted with multiplicities).
Conversely, assume that inf σess( ˜L00β) > 0. Since ˜L00β is bounded, it has a finite number m of non-positive eigenvalues
(counted with multiplicities). We denote by Θ the subspace ofH generated by the corresponding eigenvectors, and by Θ⊥ the orthogonal complement of Θ inH . There exists σ1 > 0 such that σ( ˜L00β) ∩ (0, σ1) = ∅. Then, for all
U ∈ Θ⊥, we have hL00 βU, U iH∗,H = h ˜L00βU, U iH > σ1kU k 2 H > σ1kU k 2 L2×L2.
We recall thatH is the form domain of Lβ and that Lβ is associated to the form Q, so by the form version of the
Min-Max Principle (see Theorem XIII.2 in [22]), we have inf σess(Lβ) > σ1 > 0 and Lβ has at most m non-positive
eigenvalues (counted with multiplicities).
We have thus proved (4.6) and that, in this case, the operators ˜L00β and Lβ have the same number of non-positive
Since Lβis not C-linear, it is usual to split functions into real and imaginary parts. Then, the operator Lβacting on
pairs of complex-valued functions is formally equivalent to the following operator acting on quadruplets of real-valued functions: L+β + ω2 0 0 −ω 0 L−β + ω2 ω 0 0 ω 1 0 −ω 0 0 1 , where L+βu = −u00+ (m2− ω2)u − pϕp−1 β u, (4.9) L−βu = −u00+ (m2− ω2)u − ϕp−1 β u. (4.10)
Here, L+β and L−β are R-linear operators acting on a space of real-valued functions. However, we are going to use some spectral argument which are more conveniently used with complex operators.
We denote by L2
C the usual Lebesgue space L 2
(R, C) endowed with its usual complex structure. Then we define H1
C and H 2,∗
C accordingly. We also define D β
C as Dβ, with H
2,∗replaced by H2,∗
C . Then we define the operators L + β and L−β by D(L+β) = D(L−β) = DCβ× L2 C and, for u in D β C× L 2 C, L + βu and L −
βu are defined by (4.9) and (4.10). These
are in particular C-linear operators. For λ ∈ R \ {1} we set (see Figure1)
Λ(λ) := λ + λω
2
1 − λ.
Proposition 4.5. The operators L+β and L−β are selfadjoint and bounded from below on L2
C. Moreover, for λ ∈ R\{1},
(i) λ ∈ σ(Lβ) if and only if Λ(λ) ∈ σ(L+β) ∪ σ(L − β),
(ii) we have
dim(ker(Lβ− λ)) = dim(ker(L+β − Λ(λ))) + dim(ker(L −
β − Λ(λ))), (4.11)
and in particular,
n( ˜L00β) = n(L+β) + n(L−β), (4.12)
(iii) λ ∈ σess(Lβ) if and only if Λ(λ) ∈ σess(L+β) ∪ σess(L−β).
In (4.11) and (4.12), the left-hand sides are dimensions of real vector spaces while the right-hand sides are dimensions of complex vector spaces.
Proof. • As in the proof of Proposition4.4, we can check that L+β and L−β are the selfadjoint operator corresponding to the sesquilinear forms
q+β : (u, w) 7→ hu0, w0iL2 C + (m2− ω2) hu, wi L2 C − phϕp−1β u, wiL2 C+ βu(0) ¯w(0) (4.13) and qβ−: (u, w) 7→ hu0, w0iL2 C + (m2− ω2) hu, wiL2 C − hϕp−1β u, wiL2 C + βu(0) ¯w(0), (4.14)
which are closed, symmetric and bounded from below.
• Let λ ∈ R. Let U = (u, v) and F = (f, g) in L2× L2. We write u = u
1+ iu2 where u1and u2are real valued. We
use similar notation for v, f and g. Then u belongs to Dβ if and only if u1 and u2 belong to DCβ and in this case
(Lβ− λ)U = F ⇐⇒ (L+β + ω2− λ)u1− ωv2 = f1, (L−β + ω2− λ)u 2+ ωv1 = f2, (1 − λ)v1+ ωu2 = g1, (1 − λ)v2− ωu1 = g2. If λ 6= 1 this gives (Lβ− λ)U = F ⇐⇒ (L+β − Λ(λ))u1 = f1+1−λωg2, (L−β − Λ(λ))u2 = f2−1−λωg1, v1 = g11−λ−ωu2, v2 = g21−λ+ωu1. (4.15)
• We set K+(λ) = ker(L+
β − Λ(λ)) and denote by K +
R(λ) the R-linear subspace of real-valued functions in K +(λ).
Given u ∈ Dβ
C, we have u ∈ K
+(λ) if and only if Re(u) and Im(u) are in K+
R(λ). A family of lineraly independant
vectors in K+
R(λ) is also a family of linearly independant vectors in K
+(λ), so dim R(K
+
R(λ)) 6 dimC(K
+(λ)). In
particular, if the left-hand side if infinite, then so is the right-hand side. Now assume that dimR(K+
R(λ)) if finite
(possibly 0) and consider a basis e = (e1, . . . , em) of KR+(R) (with m ∈ N). Let u ∈ K+(λ). Then Re(u) and Im(u)
belong to KR+(λ) and are R-linear combinations of vectors in e, so u is a C-linear combination of vectors in e. This proves that
dimCK+(λ) = dimRK+
R(λ).
We similarly define K−(λ) and K−
R(λ) and see that dimCK
−(λ) = dim RK
− R(λ).
If e+1, . . . , e+
m+are linearly independent vectors in K
+
R(λ) and e −
1, . . . , e−m− are linearly independent vectors in K
− R(λ)
(m± may be zero), then
e+1, iωe + 1 1 − λ , . . . , e+1,iωe + m+ 1 − λ ! , ie−1, − ωe − 1 1 − λ , . . . , ie−1, −ωe − m− 1 − λ ! (4.16) is a family of linearly independent vectors in ker(Lβ− λ), so
dim(ker(Lβ) − λ) > dim(KR+(λ)) + dim(KR−(λ)). (4.17)
In particular, if the right-hand side is infinite, then so is the left-hand side. Now assume that the right-hand side is finite. If the above families span K+
R(λ) and K −
R(λ), then (4.16) span ker(Lβ− Λ(λ)), so the inequality in (4.17) is an
equality and (4.11) is proved. Since Λ is a bijection from (−∞, 0) to itself, (4.12) follows.
• Assume that Λ(λ) ∈ ρ(L+β) ∩ ρ(L−β). Let F = (f1+ if2, g1+ ig2) ∈ L2× L2. Let (u1, u2, v1, v2) be the unique
solution of (4.15) and U = (u1+ iu2, v1+ iv2). Then U ∈ D(Lβ) and (Lβ− λ)U = F , so (Lβ− λ) is surjective. We
already know that λ is not an eigenvalue of Lβ, so (Lβ− λ) is bijective. This implies that λ is in the resolvent set of
Lβ.
Conversely, assume that λ ∈ ρ(Lβ) and let f = f1+ if2∈ L2C. We denote by u1 (resp. u2) the first component of
(Lβ− λ)−1(f1, 0, 0, 0) (resp. (Lβ− λ)−1(f2, 0, 0, 0)). Then u = u1+ iu2is such that (L+β− Λ(λ))u = f , and we deduce
that Λ(λ) ∈ ρ(L+β). Similarly, Λ(λ) ∈ ρ(L−β). This proves (i). Then (iii) follows from (i) and (ii).
Figure 1. Graph of λ 7→ Λ(λ)
Remark 4.6. If we denote by Λ− and Λ+ the restrictions of Λ to (−∞, 1) and (1, +∞), then Λ−: (−∞, 1) → R and
Λ+: (1, +∞) → R are increasing bijections. Therefore,
σess(Lβ) \ {1} = Λ−1− (σess(L+β) ∪ σess(L−β)) ∪ Λ −1
Furthermore, since σess(L+β) ∪ σess(L−β) contains a neighborhood of +∞ (see Proposition 4.7below) and σess(Lβ) is
closed, we have that 1 ∈ σess(Lβ). More precisely,
σess(Lβ) = [σ1, 1] ∪ [σ2, +∞),
with σ1= Λ−1− (m2− ω2) ∈ (0, 1) and σ2= Λ−1+ (m
2− ω2) > 1.
With Propositions 4.4 and4.5, we can deduce the spectral properties of ˜L00β from those of L+β and L−β, which we now describe.
Proposition 4.7. Let β ∈ (−β0, β0).
(i) We have σess(L+β) = σess(L−β) = [m2− ω2, +∞).
(ii) The first eigenvalue of L−β is 0, it is simple, and the corresponding eigenspace is spanned by ϕβ.
(iii) 0 is an eigenvalue of L+β if and only if β = 0. Moreover, ker(L+0) = span(∂xϕ0) and the negative spectrum of L+0
reduces to a simple eigenvalue.
Proof. It is known from [2, Theorem I-3.1.4] that the essential spectrum of −∂2
x+ (m2− ω2) is [m2− ω2, +∞). As L + β
and L−β are relatively compact perturbations of this operator, (i) follows from the Weyl Theorem (see, e.g., [15, IV-Theorem 5.35]).
Since ϕβ∈ Dβ, L−βϕβ= 0 and ϕβ > 0, the first eigenvalue of L−β is 0, it is simple and the rest of the spectrum is
positive (see, e.g., [3, Chapter 2]). This proves (ii). As for (iii), we observe that ϕβ satisfies
−ϕ00β+ (m2− ω2)ϕ
β− ϕpβ= 0 (4.18)
on (−∞, 0) and on (0, +∞). When β = 0, ϕ0 is smooth and (4.18) holds on R. After differentiation, we see that ϕ00
belongs to ker(L+0). By Theorem 3.3 in [3, Chapter 2], 0 is a simple eigenvalue of L+0 and the corresponding eigenspace is spanned by ϕ00. Moreover, by [16, Lemma 4.16], L+0 has one simple negative eigenvalue.
Now assume that β 6= 0. Let u ∈ ker(L+β). In particular, u satisfies −u00+ (m2− ω2)u − pϕp−1
β u = 0 (4.19)
on (−∞, 0) and on (0, +∞). Since ϕ0β is also a solution of (4.19) on (−∞, 0) and (0, +∞), there exist µ−, µ+ ∈ R
such that u = µ−ϕ0β on (−∞, 0) and u = µ+ϕ0β on (0, +∞). As u is continuous at 0, we have
µ−ϕ0β(0−) = u(0) = µ+ϕ0β(0
+). (4.20)
Thus µ− = −µ+ because ϕ0β(0
+) = −ϕ0
β(0−) 6= 0. Moreover, ϕβ and u both satisfy the jump condition in (4.4), so
βϕβ(0) = ϕ0β(0 + ) − ϕ0β(0−) = 2ϕ0β(0 + ) (4.21) and βu(0) = u0(0+) − u0(0−) = µ+ ϕ00β(0 +) + ϕ00 β(0 −). (4.22)
On the other hand, by (4.18),
ϕ00β(0±) = lim x→0±ϕ 00 β(x) = (m 2− ω2)ϕ β(0) − ϕ p β(0). (4.23)
Using (4.20)–(4.23) we now have
µ+β2 2 ϕβ(0) = 2µ+ (m 2− ω2)ϕ β(0) − ϕ p β(0).
Since ϕβ(0) 6= 0, it follows that
µ+ϕp−1β (0) = µ+ m2− ω2−β 2 4 . But a direct computation using (4.1) gives
ϕp−1β (0) = p + 1 2 m2− ω2−β 2 4 .
Since n(L−β) = 0, (4.12) now gives n( ˜L00β) = n(L+β). By Proposition4.7, L+0 has a simple negative eigenvalue, has 0 as a simple eigenvalue, and the rest of its spectrum is positive. We now determine the number of negative eigenvalue of L+β for all β ∈ (−β0, β0) by a perturbation argument.
Proposition 4.8. Let β ∈ (β0, β0). Then
n(L+β) = (
1 if β 6 0, 2 if β > 0.
Proof. We recall that L+β is associated to the form q+β defined in (4.13). This form is bounded from below and closed with dense domain H1
C. Moreover, for all u ∈ H 1
C, the map β 7→ q +
β(u, u) is analytic (the map β 7→ ϕβ is
pointwise analytic and locally bounded in L∞), so the family of operators L+β is analytic of type (B) in the sense of Kato (see [15, Sec. VII.4.2]). Thus, by analytic perturbation of L+0, there exist β1∈ (0, β0) and an analytic function
λ : (−β1, β1) → R such that λ(0) = 0 and, for all β ∈ (−β1, β1), λ(β) is a simple eigenvalue of L+β, L+β has a simple
eigenvalue smaller than λ(β), and the rest of its spectrum is positive. Furthermore, there exists an analytic function f : (−β1, β1) → L2C such that, for β ∈ (−β1, β1), f (β) belongs to DCβ and is an eigenfunction corresponding to the
eigenvalue λ(β).
In particular, there exist λ1∈ R and f1∈ L2Csuch that
λ(β) = βλ1+ O(β2), (4.24)
f (β) = ϕ00+ βf1+ O(β2). (4.25)
From (1.9), ϕβ is also analytic in β ∈ (−β0, β0) as a function in H1(R), so there exists g1∈ HC1 such that
ϕβ= ϕ0+ βg1+ O(β2). (4.26)
For β small, the sign of λ(β) is given by the sign of λ1, which we now compute. We have
hL+βf (β), ϕ00iL2 C = hλ(β)f (β), ϕ00iL2 C = λ1βkϕ00k 2 L2 C + O(β2). (4.27)
On the other hand, since L+β is selfadjoint and f (β), ϕ00∈ Dβ,
hL+ βf (β), ϕ 0 0iL2 C = hf (β), L+βϕ00iL2 C . Then, by (4.26), L+βϕ00= (L+β − L+ 0)ϕ00= −p(ϕ p−1 β − ϕ p−1 0 )ϕ00= −βp(p − 1)ϕ p−2 0 ϕ00g1+ O(β2).
With (4.25), this yields
hL+βf (β), ϕ00iL2 C = −βhϕ 0 0, p(p − 1)ϕ p−2 0 ϕ 0 0g1iL2 C+ O(β 2). (4.28)
A straightforward calculation using that L−βϕ0= 0 gives
p(p − 1)ϕp−20 (ϕ00)2= L+0 (m2− ω2)ϕ
0− ϕp0. (4.29)
Now consider an arbitrary ψ ∈ H1(R, R). Differentiating the identity qβ−(ψ, ϕβ) = 0 with respect to β at β = 0 yields
q+0(ψ, g1) = −ψ(0)ϕ0(0). (4.30)
In view of (4.29) and (4.30), (4.28) then becomes hL+ βf (β), ϕ 0 0iL2 C = −βq + 0 (m 2− ω2)ϕ 0− ϕp0, g1 + O(β2) = β[(m2− ω2)ϕ0(0)2− ϕ0(0)p+1] + O(β2).
Combining this with (4.27), we obtain
λ1= (m2− ω2)ϕ 0(0)2− ϕ0(0)p+1 kϕ0 0k2L2 C . But from (1.9) we have
ϕ0(0)p−1=
p + 1
2 (m
2
− ω2) > (m2− ω2),
hence λ1 < 0. It follows that there exists β2 ∈ (0, β1) such that L+β has exactly one negative eigenvalue for all
Finally, there exists κ ∈ R such that L+β > κ for all β ∈ (−β0, β0), so the negative eigenvalues of L+β are in [κ, 0).
Moreover, we know from Proposition4.7that 0 is not an eigenvalue of L+β if β 6= 0. We define the contour Γ as the boundary of the set Ω = (κ − 1, 0) + i(−1, 1). Since Γ is in the resolvent set of L+β for β 6= 0, we know from the analytic perturbation theory (see [15, VII-Section 1.3]) that the number of eigenvalues of L+β in Ω does not depend on β ∈ (−β0, 0) or β ∈ (0, β0). Since L+ω,β2 has two negative eigenvalues and L
+
ω,−β2 has exactly one, we have n(L
+ β) = 1
for all β ∈ (−β0, 0) and n(L+β) = 2 for all β ∈ (0, β0).
Combining Propositions 4.4,4.5,4.7and4.8with β = γ − αω, we finally obtain the following result. Proposition 4.9. Suppose m2− ω2> (γ − αω)2/4. Then
nω=
(
1 if γ − αω 6 0, 2 if γ − αω > 0.
Remark 4.10. For γ − αω > 0, the operator ˜L00β restricted to Hrad has only one negative eigenvalue (see [17,
Lemma 21]). Hence, dωdQ(Φω) < 0 implies orbital instability in Hrad, and so orbital instability inH .
4.2. Slope condition. We shall now turn our attention to the slope condition in order to classify various stabil-ity/instability regimes. We still consider α, γ and ω satisfying (1.8), and we now restore the dependence on ω in the notation — which was dropped in the previous subsection, where the parameter β = γ − αω played the key role.
From (1.9) and (1.11), we get Q(Φω) = −ωkϕωk2L2− α 2|ϕω(0)| 2 = −C(ω) 4ω (p − 1)√m2− ω2 Z ∞ τ (ω) sechp−14 (y) dy −α 2C(ω) 1 − (γ − αω) 2 4(m2− ω2) p−12 , (4.31) where C(ω) = (p + 1)(m 2− ω2) 2 p−12 and τ (ω) = tanh−1 −(γ − αω) 2√m2− ω2 . We first investigate the stability of standing waves when p = 3, in which case (4.31) reduces to
Q(Φω) = 2(m2− ω2) −2ω √ m2− ω2 1 + γ − αω 2√m2− ω2 −α 2 1 − (γ − αω) 2 4(m2− ω2) = −4ωpm2− ω2− 2ω(γ − αω) − α(m2 − ω2) +α 4(γ − αω) 2.
We shall inspect the derivative of Q(Φω) with respect to ω, which is given by
d dωQ(Φω) = 4ω2 √ m2− ω2 − 4 p m2− ω2+ α 3 2 + 6α ω − γ 2 +α 2 2 . (4.32)
In the following theorem we address the case when either α = 0 or γ = 0. Let us first remark that in these cases there exists an H1 solution of (1.5). Indeed, if
α = 0, |γ| < 2m and ω ∈ (−ωγ, ωγ), with ωγ= r m2−γ 2 4 , or if γ = 0 and ω ∈ (−ωα, ωα), with ωα= 2m √ 4 + α2,
then the admissibility relation (1.8) is satisfied. Theorem 4.11. Let p = 3 and m > 0.
(i) Suppose that α = 0, |γ| < 2m and |ω| 6 ωγ. Let
˜ ωγ = s 16m2− γ2+ γp γ2+ 32m2 32 = m s 1 2 + γ p γ2+ 32m2+ γ. • For γ < 0, eiωtΦ
ω is orbitally stable if |ω| > ˜ωγ and orbitally unstable if |ω| < ˜ωγ.
• For γ > 0, eiωtΦ
(ii) Suppose that γ = 0 and |α| < 2p√5 − 2. We set ω±α = m √ 2 s 1 ∓√ |κ| 4 + κ2, where κ = 1 4 α3 2 + 6α . Suppose α < 0.
• If ω ∈ (−ωα, −ωα−) then eiωtΦω is orbitally stable.
• If ω ∈ (−ω−
α, 0) then eiωtΦω is orbitally unstable.
• If ω ∈ (0, ω+
α) then eiωtΦω is orbitally unstable on Hrad.
• If ω ∈ (ω+
α, ωα) then eiωtΦω is linearly unstable.
Suppose α > 0.
• If ω ∈ (−ωα, −ωα+) then eiωtΦω is linearly unstable.
• If ω ∈ (−ω+
α, 0) then eiωtΦω is orbitally unstable onHrad.
• If ω ∈ (0, ω−
α) then eiωtΦω is orbitally unstable.
• If ω ∈ (ω−
α, ωα) then eiωtΦω is orbitally stable.
Let us remark that orbitally instability on Hrad implies orbitally instability onH .
Proof. (i) Since α = 0, we have dωdQ(Φω) = 0 if and only if
4ω2− 2m2= γpm2− ω2,
that is,
16ω4+ (γ − 16m2)ω2+ (4m4− γ2m2) = 0 and sign(2ω2− m2) = sign(γ).
The only possibility is ω2 = ˜ω2 γ. Since
d
dωQ(Φω) is negative when ω = 0 and goes to +∞ when ω goes to ±µ, we
deduce that dωdQ(Φω) < 0 if and only if |ω| 6 ˜ωγ. Furthermore, Proposition 4.9 with α = 0 yields nω = 1 if γ < 0
and nω= 2 if γ > 0. Hence, the conclusions in (i) follow from Proposition 4.1, Proposition4.2and Remark 4.10.
(ii) We now consider the case γ = 0. We have that dωdQ(Φω) = 0 if and only if
ω4− m2ω2+ m4
4 + κ2 = 0 and sign(2ω
2− m2) = sign(−κω).
The solutions are −ω−
α and ωα+. Then, for |ω| < ωα, we have dωdQ(Φω) < 0 if and only if ω ∈ (−ωα−, ω+α). Furthermore,
nω= 1 if αω > 0 and nω = 2 if αω < 0. Hence, the conclusions again follow from Proposition4.1, Proposition 4.2
and Remark4.10.
Remark 4.12. Notice that ˜ωγ > ωγ when γ > 2m/
√
3. In this case, eiωtΦ
ωis orbitally unstable for all ω ∈ (−ωγ, ωγ).
Similarly, if |α| > 2p√5 − 2 then ω+
α > ωα, so the set of ω for which we have linear instability is empty.
We next give some results with non-zero coupling constants, γ 6= 0 and α 6= 0. We first observe that the right-hand side of (4.32) vanishes for
γ = ˜γ(α, ω) := 2 4 + α2 4ω2 √ m2− ω2 − 4 p m2− ω2+ α 3 2 + 6α ω . It follows that sgn d dωQ(Φω) = − sgn(γ − ˜γ).
The following theorem is then proved using Proposition4.1, Proposition4.2and Remark4.10, similarly to the proof of Theorem4.11.
Theorem 4.13. Let p = 3 and consider ω ∈ (−m, m), α ∈ R and γ ∈ R satisfying (1.8).
(i) Suppose γ − αω < 0. Then eiωtΦω is orbitally stable if γ < ˜γ and orbitally unstable if γ > ˜γ.
(ii) Suppose γ − αω > 0. Then eiωtΦω is linearly unstable if γ < ˜γ and orbitally unstable inHrad if γ > ˜γ.
Remark 4.14. For any fixed α ∈ R, there always exist values of the parameters ω, γ satisfying the above conditions for stability/instability. For instance, conditions (1.8), γ < αω and γ < ˜γ are all satisfied provided
0 < ˜γ − αω + 2pm2− ω2= 8 4 + α2 2ω2− m2 √ m2− ω2 + 2 p m2− ω2+ 8αω 4 + α2.
We now consider more general values of the power 1 < p < 5. To keep the exposition simple enough, we focus on the cases where the coupling constants α, γ have the same sign. Of course, mixed cases could also be considered. Lemma 4.15. Let 1 < p < 5.
(i) Suppose α, γ > 0. We have
d dωQ(Φω) < 0 for ω ∈ −m 2 p p − 1, 0 and d dωQ(Φω) > 0 for ω ∈ αγ 4 + α2, αm2 γ ∩m 2 p p − 1, m, whenever these intervals are not empty.
(ii) Suppose α, γ < 0. We have d dωQ(Φω) < 0 for ω ∈ 0,m 2 p p − 1∩ αγ 4 + α2, αm2 γ and d dωQ(Φω) > 0 for ω ∈ αm2 γ , αγ 4 + α2 ∩m 2 p p − 1, m, whenever these intervals are not empty.
Proof. We only prove (i), as (ii) is proved by similar calculations. We rewrite (4.31) as Q(Φω) = C1(ω)I(ω) + C2(ω), where C1(ω) = − 4 p − 1 p + 1 2 p−12 ω(m2− ω2)p−12 − 1 2, I(ω) = Z ∞ τ (ω)
sechp−14 (y) dy,
C2(ω) = − α 2 p + 1 8 p−12 (4(m2− ω2) − (γ − αω)2)p−12 . We first find ∂C2 ∂ω = − α p − 1 p + 1 8 p−12 (4(m2− ω2) − (γ − αω)2))p−12 −1(−8ω + 2α(γ − αω)). It follows that ∂C2 ∂ω > 0 if ω > αγ 4 + α2. (4.33)
We next determine the sign of ∂ω∂ (C1I) assuming ω > 0. Since I is positive and C1 is negative for ω > 0, we will have
∂ ∂ω(C1I) = ∂C1 ∂ω I + C1 ∂I ∂ω > 0, provided ∂C1 ∂ω > 0 and ∂I
∂ω < 0. On the one hand, we have
∂C1 ∂ω = − 4 p − 1 p + 1 2 p−12 (m2− ω2)p−12 −1 2 + 8 p − 1 p + 1 2 p−12 2 p − 1− 1 2 ω2(m2− ω2)p−12 −3 2, which is positive if |ω| > m 2 p p − 1. (4.34) On the other, ∂I ∂ω = −sech 4 p−1(τ ) ∂ ∂ωτ = − 1 √ m2− ω2 1 − (γ − αω) 2 4(m2− ω2) p−12 2αm2− 2γω 4(m2− ω2) − (γ − αω)2, which is negative if ω <αm 2 γ . (4.35)
Hence, it follows from (4.33)–(4.35) that dωd Q(Φω) > 0 if ω ∈ αγ 4 + α2, αm2 γ ∩m 2 p p − 1, m. We now show that dωdQ(Φω) < 0 for ω ∈ −m2
√
p − 1, 0. As C1 and I are both positive for ω ∈ (−m, 0), we will have
∂ ∂ω(C1I) = ∂C1 ∂ω I + C1 ∂I ∂ω < 0, provided ∂C1 ∂ω < 0 and ∂I
∂ω < 0. From the previous calculations, we know that ∂C1(ω) ∂ω < 0 if |ω| <m 2 p p − 1 (4.36) and ∂ω∂I < 0 if ω <αm 2 γ . (4.37) Finally, ∂C2 ∂ω < 0 if ω < αγ 4 + α2. (4.38)
Since α > 0 and γ > 0, we conclude from (4.36)–(4.38) that dωd Q(Φω) < 0 for all ω ∈ −m2
√
p − 1, 0.
We finally combine Lemma4.15with the spectral conditions in Proposition4.9to get the following result. Theorem 4.16. Let 1 < p < 5 and m = 1.
(i) Let α, γ > 0. If 4+ααγ2 < 1 and 1 <
γ2
4+α2 +
2γ α(4+α2)
p
4 + α2− γ2, then there exists ω satisfying (1.8) and
ω ∈ αγ 4 + α2, α γ ∩ 1 2 p p − 1, 1 . For such ω, the standing wave eiωtΦ
ω is orbitally stable.
If αγ < 2p4 + α2− γ2, then there exists ω satisfying (1.8) and
ω ∈ −1 2 p p − 1, 0 . For such ω, eiωtΦ
ω is orbitally unstable.
(ii) Let α, γ < 0. If 1 < αγ, then there exists ω ∈ R satisfying (1.8) and ω ∈ α γ, αγ 4 + α2 ∩ 1 2 p p − 1, 1 . For such ω, eiωtΦ
ω is orbitally stable.
If αγ < 1 2
√
p − 1, then there exists ω ∈ R satisfying (1.8) and ω ∈ 0,1 2 p p − 1 ∩ αγ 4 + α2, α γ . For such ω, eiωtΦω is orbitally unstable.
Proof. We only prove (i), the proof of (ii) being similar. The hypotheses m = 1 and γ < α imply that there exists ω ∈ R satisfying ω ∈ αγ 4 + α2, α γ ∩ 1 2 p p − 1, 1 . In particular, since 4+ααγ2 < γ α, we also have ω < γ α. Furthermore, if 1 < γ2 4+α2+ 2γ α(4+α2) p 4 + α2− γ2, then ω satisfies
the admissibility condition (1.8). Orbital stability then follows from Proposition4.9and Lemma 4.15. The condition αγ < 2p4 + α2− γ2 implies that there exists a ω ∈ −1
2
√
p − 1, 0 satisfying (1.8). Orbital
insta-bility follows from Proposition4.9, Remark 4.10and Lemma 4.15.
Remark 4.17. The condition 1 <4+αγ22 +
2γ α(4+α2)
p
